Power of approximate exponential of operator sum Say $A$ and $B$ are two non-commuting operators in an Hilbert space (think matrices, for instance). Then, one way we can approximate $C=e^{A+B}$ is as $\widetilde{C}=e^{A/2}e^{B/2}e^{A/2}e^{B/2}\simeq C$.
Question: Does it hold that $\widetilde{C}^{2}=e^{A}e^{B}e^{A}e^{B}$? It doesn't, right?
 A: Your question is a bit open ended and dependent on the matrix norms involved, etc... Purely formally, if you introduce a parameter t, something tells me you are interested in the Zassenhaus formula,
$$
e^{t(A+B)}= e^{tA}~ e^{tB} ~e^{-\frac{t^2}{2} [A,B]} \\ \times ~ 
e^{\frac{t^3}{6}(2[B,[A,B]]+ [A,[A,B]] )}   ~
e^{\frac{-t^4}{24}([[[A,B],A],A] + 3[[[A,B],A],B] + 3[[[A,B],B],B]) } \cdots
$$
Anyway, the expression you write is, by straightforward application of the CBH formula
$$
\widetilde{C}=e^{tA/2}e^{tB/2}e^{tA/2}e^{tB/2}=  (e^{tA/2}e^{tB/2})^2\\
= ( e^{\frac{t}{2}(A+B)+\frac{t^2}{8}[A,B]+ O(t^3)  } )^2 \\ 
 =e^{t(A+B)+ \frac{t^2}{4}[A,B]+ O(t^3) },
$$
which does not look like a good "approximation". The last step follows from the fact that the commutator of the exponent with itself upon squaring vanishes, so the squaring act has a degenerate, terminating CBH expansion!!
You could easily eliminate the $O(t^2)$ terms in the exponent by considering
$$
e^{tA/2}e^{tB}e^{tA/2}  =e^{t(A+B)+  O(t^3) },
$$
instead.
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